Iterating over all (i,j)-elements in a circle - c++

Question

Given two (very simplified) classes:
class Rectangle {
public:
int iL,jL; // lower left point of rectangle
int width, length; // width: in i-direction, length: in j-direction
};
class Circle {
public:
int iC,jC; // center-point of circle
int radius;
};
If I want to iterate over all elements in a Rectangle, I can simply do that by:
for (int i = iL; i < iL-width; i--)
for (int j = jL; j < jL+length; j++)
doSomething();
My problem is to implement a smart way of iterating over all elements in Circle. My current solution looks as follows:
for (int i = iC-radius; i <= iC+radius; i++)
for (int j = jC-radius; j <= jC+radius; j++)
if ( sqrt(pow(i-iC,2)+pow(j-jC,2)) <= r ) // checking if (i,j) lies within the circle (or its boundary)
doSomething();
However, for radius getting large my current solution is very time-expensive (since I touch many elements which aren't in the Circle and since I always need to evaluate pow). Can you think of a more intelligent and efficient way of iteration over all Circle-elements?

Answer 1

For every line find the first column that belongs to the circle, and walk from this column to one mirrored relative to the circle center. Pseudo code
for (int iy = - radius to radius; iy++)
dx = (int) sqrt(radius * radius - iy * iy)
for (int ix = - dx to dx; ix++)
doSomething(CX + ix, CY + iy);

Answer 2

Let the radius of the circle be r. Consider a Square of size (2r+1)*(2r+1) around the circle to be drawn. So equidistant points in square are stored in 2D array .
Now walk through every point inside the square. For every point (x,y), if (x, y) lies inside the circle (or x^2+ y^2 < r^2), then print it. For example, equidistant points form a 10x10 array, and a chosen "center" of the array at (x,y):
for i from 0 to 9 {
for j from 0 to 9 {
a = i - x
b = j - y
if a*a + b*b <= r*r {
// Do something here
}
}
}